CodeForces612AThe Text Splitting（模拟，暴力枚举）

Description
You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input
The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output
If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Sample Input
Input
5 2 3
Hello
Output
2
He
llo
Input
10 9 5
Codeforces
Output
2
Codef
orces
Input
6 4 5
Privet
Output
-1
Input
8 1 1
abacabac
Output
8
a
b
a
c
a
b
a

c

代码：

#include<stdio.h>#include<string.h>int main(){char a[101];int n,p,q;while(scanf("%d %d %d",&n,&p,&q)!=EOF){scanf("%s",a);int s=0;for(int i=0;i<=n;i++){if(s!=0)break;for(int j=0;j<=n;j++){if(i*p+j*q>n)break;if(i*p+j*q==n){printf("%d\n",i+j);s++;int l=0;for(int k=0;k<i*p;k++){printf("%c",a[l++]);if((k+1)%p==0)printf("\n");}for(int h=0;h<j*q;h++){printf("%c",a[l++]);if((h+1)%q==0)printf("\n");}break;}}}if(s==0)printf("-1\n");}return 0;}

思路：这道水题竟然坑了我一中午！！！开始写了一个代码没有考略完所有的情况wa了，后来这个程序测数据都对，就是在第一题就wa，看了好久好久好久才发现自己改的时候不小心把共有几行给删了。我…………。这道题思路是给你一个数，判断能不能用给的另外两个数组成它。如果可以输出共有几行数并且按行输出字符。两个for循环判断能否组成，可以的话if(i*p+j*q==n) 按行输出，注意换行。不可以就输出-1.